If x=e^{theta} sin theta and y=e^{theta} cos | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given,$x=e^{	heta} \sin 	heta$ and $y=e^{	heta} \cos 	heta$. We differentiate both with respect to $	heta$: $\frac{dx}{d	heta} = e^{	heta}

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given,$x=e^{	heta} \sin 	heta$ and $y=e^{	heta} \cos 	heta$. We differentiate both with respect to $	heta$: $\frac{dx}{d	heta} = e^{	heta} \sin 	heta + e^{	heta} \cos 	heta = e^{	heta}(\sin 	heta + \cos 	heta)$ $\frac{dy}{d	heta} = e^{	heta} \cos 	heta - e^{	heta} \sin 	heta = e^{	heta}(\cos 	heta - \sin 	heta)$ Now,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{e^{	heta}(\cos 	heta - \sin 	heta)}{e^{	heta}(\cos 	heta + \sin 	heta)} = \frac{\cos 	heta - \sin 	heta}{\cos 	heta + \sin 	heta}$. Dividing numerator and denominator by $\cos 	heta$,we get $\frac{dy}{dx} = \frac{1 - 	an 	heta}{1 + 	an 	heta}$. At $(x, y) = (1, 1)$,we have $x = e^{	heta} \sin 	heta = 1$ and $y = e^{	heta} \cos 	heta = 1$. Dividing these,$\frac{y}{x} = \frac{e^{	heta} \cos 	heta}{e^{	heta} \sin 	heta} = \cot 	heta = 1$,so $	an 	heta = 1$. Substituting $	an 	heta = 1$ into the expression for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0$.

If $x=e^{\theta} \sin \theta$ and $y=e^{\theta} \cos \theta$,where $\theta$ is a parameter,then $\frac{dy}{dx}$ at $(1,1)$ is equal to

Similar Questions

$x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right), y=\sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right) \Rightarrow \frac{d y}{d x}$ is equal to

If $x = \frac{t^2}{1+t^5}$ and $y = \frac{2t^3}{1+t^5}$ where $t \neq -1$ is a parameter,then find $\frac{dy}{dx}$.

If $\sqrt{y-\sqrt{y-\sqrt{y-\ldots \infty}}} = \sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}}$,then $\frac{dy}{dx} = $

If $\tan y = \frac{2t}{1 - t^2}$ and $\sin x = \frac{2t}{1 + t^2},$ then $\frac{dy}{dx} = $

If $x = \frac{9t^2}{1+t^4}$ and $y = \frac{16t^2}{1-t^4}$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module